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A particle is moving on a circular path ...

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will be

A

`3.2ms^(-2)`

B

`2ms^(-2)`

C

`2.5ms^(-2)`

D

`4.3ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Magnitude of the net acceleration `a=sqrt(a_t^2+a_n^2)`
where, `a_t "rate of change of speed "=2 ms ^(-2)`
`a_n=(v^2)/(R)-(5)^2/(10)=2.5rArr a=sqrt((2)^2+(2.5)^2)=3.2ms^(-2)`
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Knowledge Check

  • A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 ms^(-1) and the speed is increasing at a rate of 2 m s^2 . The magnitude of net acceleration at this instant is

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